X-Git-Url: http://www.chiark.greenend.org.uk/ucgi/~ian/git?a=blobdiff_plain;f=article.tex;h=f5c366fc156132f17b8a2efe43cd7eed650a6911;hb=adb68c7bad8cd24c8a4b2ebad937b544cdfa37c6;hp=3d331786a158e419dd13ee583f7446e8859b1b71;hpb=d380695f58ab5d366203d91873ffa17ba809447e;p=topbloke-formulae.git

diff --git a/article.tex b/article.tex
index 3d33178..f5c366f 100644
--- a/article.tex
+++ b/article.tex
@@ -59,6 +59,8 @@
     {\hbox{\scriptsize$\forall$}}}%
 }
 
+\newcommand{\Largeexists}{\mathop{\hbox{\Large$\exists$}}}
+\newcommand{\Largenexists}{\mathop{\hbox{\Large$\nexists$}}}
 
 \newcommand{\qed}{\square}
 \newcommand{\proof}[1]{{\it Proof.} #1 $\qed$}
@@ -114,7 +116,7 @@ which are in $\set P$.
 \item[ $ \pendsof{C}{\set P} $ ]
 $ \{ E \; | \; E \in \pancsof{C}{\set P}
   \land \mathop{\not\exists}_{A \in \pancsof{C}{\set P}}
-  A \neq E \land E \le A \} $ 
+  E \neq A \land E \le A \} $ 
 i.e. all $\le$-maximal commits in $\pancsof{C}{\set P}$.
 
 \item[ $ \baseof{C} $ ]
@@ -217,6 +219,17 @@ in which case we repeat for $A'$.  Since there are finitely many
 commits, this terminates with $A'' \in \pends()$, ie $A'' \le M$
 by the LHS.  And $A \le A''$.
 }
+\[ \eqn{Calculation Of Ends:}{
+  \bigforall_{C \hasparents \set A}
+    \pendsof{C}{\set P} =
+       \Bigl\{ E \Big|
+           \Bigl[ \Largeexists_{A \in \set A} 
+                       E \in \pendsof{A}{\set P} \Bigr] \land
+           \Bigl[ \Largenexists_{B \in \set A} 
+                       E \neq B \land E \le B \Bigr]
+       \Bigr\}
+}\]
+XXX proof TBD.
 
 \section{Commit annotation}
 
@@ -370,4 +383,52 @@ R$ so $D \le C$.  OK.
 
 $\qed$
 
+\subsection{Unique Base}
+
+Need to consider only $C \in \py$, ie $L \in \py$,
+and calculate $\pendsof{C}{\pn}$.  So we will consider some
+putative ancestor $A \in \pn$ and see whether $A \le C$.
+
+$A \le C \equiv A \le L \lor A \le R \lor A = C$.
+But $C \in py$ and $A \in \pn$ so $A \neq C$.  
+Thus $A \le L \lor A \le R$.
+
+By Unique Base of L and Transitive Ancestors,
+$A \le L \equiv A \le \baseof{L}$.
+
+\subsubsection{For $R \in \py$:}
+
+By Unique Base of $R$ and Transitive Ancestors,
+$A \le R \equiv A \le \baseof{R}$.
+
+But by Tip Merge condition on $\baseof{R}$,
+$A \le \baseof{L} \implies A \le \baseof{R}$, so
+$A \le \baseof{R} \lor A \le \baseof{R} \equiv A \le \baseof{R}$.
+Thus $A \le C \equiv A \le \baseof{R}$.  Ie, $\baseof{C} =
+\baseof{R}$.
+
+UP TO HERE
+
+By Tip Merge, $A \le $
+
+Let $S =
+   \begin{cases} 
+     R \in \py : & \baseof{R} \\
+     R \in \pn : & R
+   \end{cases}$.  
+Then by Tip Merge $S \ge \baseof{L}$, and $R \ge S$ so $C \ge S$.
+   
+Consider some $A \in \pn$.  If $A \le S$ then $A \le C$.
+If $A \not\le S$ then 
+
+Let $A \in \pends{C}{\pn}$.  
+Then by Calculation Of Ends $A \in \pendsof{L,\pn} \lor A \in
+\pendsof{R,\pn}$.
+
+
+
+%$\pends{C,
+
+%%\subsubsection{For $R \in \py$:}
+
 \end{document}